1. RECURSION Lecture 7 CS2110 – Fall 2015

2. Overview references to sections in text 2  Note: We’ve covered everything in JavaSummary.pptx!  What is recursion? 7.1-7.39 slide 1-7  Base case 7.1-7.10 slide 13  How Java stack frames work 7.8-7.10 slide 28-32

3. Function equals 3 public class Object { /** Return true iff this object is * the same as ob */ public boolean equals(Object b) { return this == b; } x.equals(y) is same as x == y except when x is null! y ? Object x ? This gives a null-pointer exception: null.equals(y) a0 Object equals(Object)

4. Overriding function equals 4 Override function equals in a class to give meaning to: “these two (possibly different) objects of the class have the same values in some of their fields” For those who are mathematically inclined, like any equality function, equals should be reflexive, symmetric, and transitive. Reflexive: b.equals(b) Symmetric: b.equals(c) = c.equals(b) Transitive: if b.equals(c) and c.equals(d), then b.equals(d)

5. Function equals in class Animal 5 Animal a0 name age Animal(String, int) equals() toString() … Object equals(Object) public class Animal { /** = “h is an Animal with the same values in its fields as this Animal” */ public boolean equals (Object h) { if (!(h instanceof Animal)) return false; Animal ob= (Animal) h; return name.equals(ob.name) && age == ob.age; } toString() 1. Because of h is an Animal in spec, need the test h instanceof Animal

6. Function equals in class Animal 6 Animal a0 name age Animal(String, int) equals() toString() … Object equals(Object) public class Animal { /** = “h is an Animal with the same values in its fields as this Animal” */ public boolean equals (Object h) { if (!(h instanceof Animal)) return false; Animal ob= (Animal) h; return name.equals(ob.name) && age == ob.age; } toString() 2. In order to be able to reference fields in partition Animal, need to cast h to Animal

7. Function equals in class Animal 7 Animal a0 name age Animal(String, int) equals() toString() … Object equals(Object) public class Animal { /** = “h is an Animal with the same values in its fields as this Animal” */ public boolean equals (Object h) { if (!(h instanceof Animal)) return false; Animal ob= (Animal) h; return name.equals(ob.name) && age == ob.age; } toString() 3. Use String equals function to check for equality of String values. Use == for primitive types

8. Why can’t the parameter type be Animal? 8 Animal a0 name age Animal(String, int) equals() toString() … Object equals(Object) public class Animal { /** = “h is an Animal with the same values in its fields as this Animal” */ public boolean equals (Animal h) { if (!(h instanceof Animal)) return false; Animal ob= (Animal) h; return name.equals(ob.name) && age == ob.age; } toString() What is wrong with this?

9. Sum the digits in a non-negative integer 9 E.g. sum(7) = 7 /** return sum of digits in n. * Precondition: n >= 0 */ public static int sum(int n) { if (n < 10) return n; // { n has at least two digits } // return first digit + sum of rest return sum(n/10) + n%10 ; } sum calls itself! E.g. sum(8703) = sum(870) + 3;

10. Two issues with recursion 10 1. Why does it work? How does the method executed? /** return sum of digits in n. * Precondition: n >= 0 */ public static int sum(int n) { if (n < 10) return n; // { n has at least two digits } // return first digit + sum of rest return sum(n/10) + n%10 + ; } sum calls itself! 2. How do we understand a given recursive method or how do we write/develop a recursive method?

11. Stacks and Queues 11 top element 2nd element... bottom element stack grows Stack: list with (at least) two basic ops: * Push an element onto its top * Pop (remove) top element Last-In-First-Out (LIFO) Like a stack of trays in a cafeteria first second … last Queue: list with (at least) two basic ops: * Append an element * Remove first element First-In-First-Out (FIFO) Americans wait in a line the Brits wait in a queue !

12. local variables parameters return info Stack Frame 12 a frame A “frame” contains information about a method call: At runtime Java maintains a stack that contains frames for all method calls that are being executed but have not completed. Method call: push a frame for call on stack assign argument values to parameters execute method body. Use the frame for the call to reference local variables parameters. End of method call: pop its frame from the stack; if it is a function leave the return value on top of stack.

13. Frames for methods sum main method in the system 13 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); } frame: n ___ return info frame: r ___ args ___ return info frame: ? return info Frame for method in the system that calls method main

14. Example: Sum the digits in a non-negative integer 14 ? return info Frame for method in the system that calls method main: main is then called system r ___ args ___ return info main public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

15. Example: Sum the digits in a non-negative integer 15 ? return info Method main calls sum: system r ___ args ___ return info main n ___ return info 824 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

16. Example: Sum the digits in a non-negative integer 16 ? return info n >= 10 sum calls sum: system r ___ args ___ return info main n ___ return info 824 n ___ return info 82 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

17. Example: Sum the digits in a non-negative integer 17 ? return info n >= 10. sum calls sum: system r ___ args ___ return info main n ___ return info 824 n ___ return info 82 n ___ return info 8 10 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

18. Example: Sum the digits in a non-negative integer 18 ? return info n < 10 sum stops: frame is popped and n is put on stack: system r ___ args ___ return info main n ___ return info 824 n ___ return info 82 n ___ return info 8 8 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

19. Example: Sum the digits in a non-negative integer 19 ? return info Using return value 8 stack computes 8 + 2 = 10 pops frame from stack puts return value 10 on stack r ___ args ___ return info main n ___ return info 824 n ___ return info 82 8 10 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

20. Example: Sum the digits in a non-negative integer 20 ? return info Using return value 10 stack computes 10 + 4 = 14 pops frame from stack puts return value 14 on stack r ___ args ___ return info main n ___ return info 824 10 14 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

21. Example: Sum the digits in a non-negative integer 21 ? return info Using return value 14 main stores 14 in r and removes 14 from stack r ___ args __ return info main 14 public static int sum(int n) { if (n < 10) return n; return sum(n/10) + n%10; } public static void main( String[] args) { int r= sum(824); System.out.println(r); }

22. Summary of method call execution 22 Memorize this!  1. A frame for a call contains parameters local variables and other information needed to properly execute a method call.  2. To execute a method call: push a frame for the call on the stack, assign values to parameters, execute method body, pop frame for call from stack, and (for a function) push returned value on stack When executing method body look in frame for call for parameters and local variables.

23. Questions about local variables 23 public static void m(…) { … while (…) { int d= 5; … } In a call m() when is local variable d created and when is it destroyed? Which version of procedure m do you like better? Why? public static void m(…) { int d; … while (…) { d= 5; … }

24. Recursion is used extensively in math 24 Math definition of n factorial E.g. 3! = 3*2*1 = 6 0! = 1 n! = n * (n-1)! for n > 0 Math definition of b c for c >= 0 b 0 = 1 b c = b * b c-1 for c > 0 Easy to make math definition into a Java function! public static int fact(int n) { if (n == 0) return 1; return n * fact(n-1); } Lots of things defined recursively: expression grammars trees …. We will see such things later

25. Two views of recursive methods 25  How are calls on recursive methods executed? We saw that. Use this only to gain understanding / assurance that recursion works  How do we understand a recursive method — know that it satisfies its specification? How do we write a recursive method? This requires a totally different approach. Thinking about how the method gets executed will confuse you completely! We now introduce this approach.

26. Understanding a recursive method 26 Step 1. Have a precise spec! /** = sum of digits of n. * Precondition: n >= 0 */ public static int sum(int n) { if (n < 10) return n; // n has at least two digits return sum(n/10) + n%10 ; } Step 2. Check that the method works in the base case(s): Cases where the parameter is small enough that the result can be computed simply and without recursive calls. If n < 10 then n consists of a single digit. Looking at the spec we see that that digit is the required sum.

27. Step 3. Look at the recursive case(s). In your mind replace each recursive call by what it does according to the method spec and verify that the correct result is then obtained. return sum(n/10) + n%10; Understanding a recursive method 27 Step 1. Have a precise spec! /** = sum of digits of n. * Precondition: n >= 0 */ public static int sum(int n) { if (n < 10) return n; // n has at least two digits return sum(n/10) + n%10 ; } Step 2. Check that the method works in the base case(s). return (sum of digits of n/10) + n%10; // e.g. n = 843

28. Step 3. Look at the recursive case(s). In your mind replace each recursive call by what it does acc. to the spec and verify correctness. Understanding a recursive method 28 Step 1. Have a precise spec! /** = sum of digits of n. * Precondition: n >= 0 */ public static int sum(int n) { if (n < 10) return n; // n has at least two digits return sum(n/10) + n%10 ; } Step 2. Check that the method works in the base case(s). Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the pars of the method. n/10 < n

29. Step 3. Look at the recursive case(s). In your mind replace each recursive call by what it does according to the spec and verify correctness. Understanding a recursive method 29 Step 1. Have a precise spec! Step 2. Check that the method works in the base case(s). Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the pars of the method Important! Can’t do step 3 without it Once you get the hang of it this is what makes recursion easy! This way of thinking is based on math induction which we will see later in the course.

30. Step 3. Look at all other cases. See how to define these cases in terms of smaller problems of the same kind. Then implement those definitions using recursive calls for those smaller problems of the same kind. Done suitably point 4 is automatically satisfied. Writing a recursive method 30 Step 1. Have a precise spec! Step 2. Write the base case(s): Cases in which no recursive calls are needed Generally for “small” values of the parameters. Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the pars of the method

31. Step 3. Look at all other cases. See how to define these cases in terms of smaller problems of the same kind. Then implement those definitions using recursive calls for those smaller problems of the same kind. Examples of writing recursive functions 31 Step 1. Have a precise spec! Step 2. Write the base case(s). For the rest of the class we demo writing recursive functions using the approach outlined below. The java file we develop will be placed on the course webpage some time after the lecture.

32. The Fibonacci Function 32 Mathematical definition: fib(0) = 0 fib(1) = 1 fib(n) = fib(n  1) + fib(n  2) n ≥ 2 Fibonacci sequence: 0 1 1 2 3 5 8 13 … /** = fibonacci(n). Pre: n >= 0 */ static int fib(int n) { if (n <= 1) return n; // { 1 < n } return fib(n-2) + fib(n-1); } two base cases! Fibonacci (Leonardo Pisano) 1170-1240? Statue in Pisa Italy Giovanni Paganucci 1863

33. Check palindrome-hood 33 A String palindrome is a String that reads the same backward and forward. A String with at least two characters is a palindrome if  (0) its first and last characters are equal and  (1) chars between first & last form a palindrome: e.g. AMANAPLANACANALPANAMA A recursive definition! have to be the same have to be a palindrome

34. Example: Is a string a palindrome? 34 isPal(“racecar”) returns true isPal(“pumpkin”) returns false /** = "s is a palindrome" */ public static boolean isPal(String s) { if (s.length() <= 1) return true; // { s has at least 2 chars } int n= s.length()-1; return s.charAt(0) == s.charAt(n) && isPal(s.substring(1,n)); } Substring from s[1] to s[n-1]

35. 35  A man a plan a caret a ban a myriad a sum a lac a liar a hoop a pint a catalpa a gas an oil a bird a yell a vat a caw a pax a wag a tax a nay a ram a cap a yam a gay a tsar a wall a car a luger a ward a bin a woman a vassal a wolf a tuna a nit a pall a fret a watt a bay a daub a tan a cab a datum a gall a hat a fag a zap a say a jaw a lay a wet a gallop a tug a trot a trap a tram a torr a caper a top a tonk a toll a ball a fair a sax a minim a tenor a bass a passer a capital a rut an amen a ted a cabal a tang a sun an ass a maw a sag a jam a dam a sub a salt an axon a sail an ad a wadi a radian a room a rood a rip a tad a pariah a revel a reel a reed a pool a plug a pin a peek a parabola a dog a pat a cud a nu a fan a pal a rum a nod an eta a lag an eel a batik a mug a mot a nap a maxim a mood a leek a grub a gob a gel a drab a citadel a total a cedar a tap a gag a rat a manor a bar a gal a cola a pap a yaw a tab a raj a gab a nag a pagan a bag a jar a bat a way a papa a local a gar a baron a mat a rag a gap a tar a decal a tot a led a tic a bard a leg a bog a burg a keel a doom a mix a map an atom a gum a kit a baleen a gala a ten a don a mural a pan a faun a ducat a pagoda a lob a rap a keep a nip a gulp a loop a deer a leer a lever a hair a pad a tapir a door a moor an aid a raid a wad an alias an ox an atlas a bus a madam a jag a saw a mass an anus a gnat a lab a cadet an em a natural a tip a caress a pass a baronet a minimax a sari a fall a ballot a knot a pot a rep a carrot a mart a part a tort a gut a poll a gateway a law a jay a sap a zag a fat a hall a gamut a dab a can a tabu a day a batt a waterfall a patina a nut a flow a lass a van a mow a nib a draw a regular a call a war a stay a gam a yap a cam a ray an ax a tag a wax a paw a cat a valley a drib a lion a saga a plat a catnip a pooh a rail a calamus a dairyman a bater a canal Panama

36. Example: Count the e’s in a string 36  countEm(‘e’ “it is easy to see that this has many e’s”) = 4  countEm(‘e’ “Mississippi”) = 0 /** = number of times c occurs in s */ public static int countEm(char c String s) { if (s.length() == 0) return 0; // { s has at least 1 character } if (s.charAt(0) != c) return countEm(c s.substring(1)); // { first character of s is c} return 1 + countEm (c s.substring(1)); } substring s[1..] i.e. s[1] … s(s.length()-1)